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New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

We know

Energy on any orbit is given by,

E n = 1 3 . 6 n 2 z 2

  E 1 = 1 3 . 6 1 2 * 9 [n1 = 1] (1)

For 3rd orbit

E 3 = 1 3 . 6 9 *   [n2 = 3]

E3 = 13.6ev

Δ E = E 3 E 1

= 13.6 – (13.6 * 9)

Δ E = 8 * 1 3 . 6 e v = p h o t o n e n e r g y

8 * 1 3 . 6 e v = h c λ

8 * 1 3 . 6 e v = 1 2 4 2 e v λ n m

λ = 1 1 . 4 1 5 * 1 0 9

= 114.15 * 1010m

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Given absorb energy = 10.2 eV

E f = 1 3 . 6 n 2 = 3 . 4

n 2 = 4

n = 2

 

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

mvr = n h 2 π

So momentum (mv) = n h 2 π r

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

According to definition of wave number, we can write

1λ=R (1121n2)11n2=1λR

1 n 2 = 1 1 λ R = λ R 1 λ R n = λ R λ R 1

New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

According to definition of atomic energy, we can write

E1=E0 (122132)=5E036, andE2=E0 (12212)

E1E2=59xx+4=55+4x=5

New answer posted

2 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

Based on theory 

New answer posted

2 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

According to Rutherford, e- revolves around in nucleus in circular orbit. Thus e- is always accelerating (centripetal acceleration). An accelerating change emits EM radiation and thus e- should loose energy and finally should collapse in the nucleus.

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